Problem: The equation of a circle $C$ is $x^2+y^2-14x-8y+40 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-14x) + (y^2-8y) = -40$ $(x^2-14x+49) + (y^2-8y+16) = -40 + 49 + 16$ $(x-7)^{2} + (y-4)^{2} = 25 = 5^2$ Thus, $(h, k) = (7, 4)$ and $r = 5$.